Chem Differential Eq HW Solutions Fall 2011 53

Chem Differential Eq HW Solutions Fall 2011 53 - cosh m...

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Section 3.10 Neumann and Robin Conditions 53 Thus B 0 = 0 and B m = 0 for all m 1 and so A 0 + ± m =1 A m cos a x cosh[ a ( b - y )] . Now, using u ( x, 0) = g ( x ), we Fnd g ( x )= A 0 + ± m =1 A m cosh[ a b ] cos a x. Recognizing this as a cosine series, we conclude that A 0 = 1 a ² a 0 g ( x ) dx and A m cosh[ a b ]= 2 a ² a 0 g ( x ) cos a xdx ; equivalently, for m 1, A m = 2 a cosh[ a b ] ² a 0 g ( x ) cos a x dx. 9. We follow the solution in Example 3. We have u ( x, y )= u 1 ( x, y )+ u 2 ( x, y ) , where u 1 ( x, y )= ± m =1 B m sin mx sinh my, with B m = 2 πm cosh( ) ² π 0 sin mx dx = 2 πm 2 cosh( ) (1 - ( - 1) m ); and u 2 ( x, y )= ± m =1 A m sin mx
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Unformatted text preview: cosh[ m ( π-y )] , with A m = 2 π cosh( mπ ) ² π sin mx dx = 2 πm cosh( mπ ) (1-(-1) m ) . Hence u ( x, y ) = 2 π ∞ ± m =1 (1-(-1) m ) m cosh( mπ ) sin mx ³ sinh my m + cosh[ m ( π-y )] ´ = 4 π ∞ ± k =0 sin(2 k + 1) x (2 k + 1) cosh[(2 k + 1) π ] ³ sinh[(2 k + 1) y ] (2 k + 1) + cosh[(2 k + 1)( π-y )] ´ ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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